Refrigeration Cycle: A refrigerator operates on the ideal vapor-compression refrigeration cycle and use R-134a as the working fluid. The condenser operates 1.4 MPa and the evaporator at 0.140 MPa. (a) Show the cycle on a T-s diagram, indicating the enthalpies at the end of each process, with respect to saturation lines. For a mass flow rate of 0.100 kg/s, determine (b) the net power input to the compressor, in kW, (c) the heat rejection to the environment, in kW, (d) the heat rejection from the cold refrigerated space to the evaporator, in kW, and (e) the coefficient of performance. use the formula in the picture Formula and Property Tables:QL = m(h1 – h4)Win = m(h2 – h)QH = m(h2 – h3)h1 - h4COP =h2 – hСОРinSaturated R134a - Pressure TableP, kPaTat °CSpecific EnthalpySpecific Entropyhe, kJ/kgh kJ/kgh kJ/kgSp. kJ/kg KStg. kJ/kg.K$. kJ/kg K120- 22.3222.49214.48236.970.092750.855030.94779140- 18.7727.08212.08239.160.110870.833680.94456160- 15.6031.21209.90241.110.126930814960.94190120046.29117.77156.10273.870.424410.488630.91303140052.40127.77148.90276.120.453150.457340.91050160057.88135.93141.93277.860.479110.428730.90784Superheated R134aP= 0.140 MPa, (Tat=-18.77°C)Superheated RI34aP= 1.40 MPa, (Tat = 57.88°C)T,uhT,uh°Cn/kgkJ/kgkJ/kgkJ/kg.K°Cm/kgkJ/kgkJ/kgkJ/kg.KSat0.14014215.54239.160.9446Sat0.01212258.47277.860.9078- 20600.01237260.89280.690.9163- 100.14605225.91246.360.9724700.01343271.76293.250.95350.0.15263233.23254.601.0031800.01436282.09305.070.9875

Answer: ◆ The net power input to the compressor is 5.107155 kW. ◆ The heat rejection to the…

Answered: A refrigerator operates on the ideal…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Refrigeration Cycle: A refrigerator operates on the ideal vapor-compression refrigeration cycle and use R-134a as the working fluid. The condenser operates 1.4 MPa and the evaporator at 0.140 MPa. (a) Show the cycle on a T-s diagram, indicating the enthalpies at the end of each process, with respect to saturation lines. For a mass flow rate of 0.100 kg/s, determine (b) the net power input to the compressor, in kW, (c) the heat rejection to the environment, in kW, (d) the heat rejection from the cold refrigerated space to the evaporator, in kW, and (e) the coefficient of performance.

use the formula in the picture

Formula and Property Tables:
QL = m(h1 – h4)
Win = m(h2 – h)
QH = m(h2 – h3)
h1 - h4
COP =
h2 – h
СОР
in
Saturated R134a - Pressure Table
P, kPa
Tat °C
Specific Enthalpy
Specific Entropy
he, kJ/kg
h kJ/kg
h kJ/kg
Sp. kJ/kg K
Stg. kJ/kg.K
$. kJ/kg K
120
- 22.32
22.49
214.48
236.97
0.09275
0.85503
0.94779
140
- 18.77
27.08
212.08
239.16
0.11087
0.83368
0.94456
160
- 15.60
31.21
209.90
241.11
0.12693
081496
0.94190
1200
46.29
117.77
156.10
273.87
0.42441
0.48863
0.91303
1400
52.40
127.77
148.90
276.12
0.45315
0.45734
0.91050
1600
57.88
135.93
141.93
277.86
0.47911
0.42873
0.90784
Superheated R134a
P= 0.140 MPa, (Tat=-18.77°C)
Superheated RI34a
P= 1.40 MPa, (Tat = 57.88°C)
T,
u
h
T,
u
h
°C
n/kg
kJ/kg
kJ/kg
kJ/kg.K
°C
m/kg
kJ/kg
kJ/kg
kJ/kg.K
Sat
0.14014
215.54
239.16
0.9446
Sat
0.01212
258.47
277.86
0.9078
- 20
60
0.01237
260.89
280.69
0.9163
- 10
0.14605
225.91
246.36
0.9724
70
0.01343
271.76
293.25
0.9535
0.
0.15263
233.23
254.60
1.0031
80
0.01436
282.09
305.07
0.9875

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